1. Problem Analysis

We are given two point charges \( q_1 = 4.0 \, \text{nC} \) and \( q_2 = 1.0 \, \text{nC} \). The problem statement and the accompanying figure describe a specific equipotential curve that intersects itself. This intersection occurs at the equilibrium point (null point) where the net electric field is zero.

Given:

• \( q_1 = 4.0 \times 10^{-9} \, \text{C} \)
• \( q_2 = 1.0 \times 10^{-9} \, \text{C} \)
• Potential at the null point \( V = 900 \, \text{V} \)

Goal: Find the separation distance \( d \) between the charges.

2. Locating the Null Point

Let the separation between the charges be \( d \). The null point lies on the line joining the charges, between them. Let its distance from \( q_1 \) be \( x \). Consequently, its distance from \( q_2 \) is \( d – x \).

At the null point, the electric fields cancel out:

$$ E_1 = E_2 \implies \frac{k q_1}{x^2} = \frac{k q_2}{(d-x)^2} $$

Taking the square root of both sides:

$$ \frac{\sqrt{q_1}}{x} = \frac{\sqrt{q_2}}{d-x} $$

Substitute \( q_1 = 4 \) and \( q_2 = 1 \) (units cancel):

$$ \frac{2}{x} = \frac{1}{d-x} \implies 2(d-x) = x \implies 2d = 3x \implies x = \frac{2d}{3} $$

So, the null point is at a distance \( \frac{2d}{3} \) from \( q_1 \) and \( \frac{d}{3} \) from \( q_2 \).

3. Calculating the Potential

The total electric potential \( V \) at this point is the sum of potentials due to each charge:

$$ V = \frac{k q_1}{x} + \frac{k q_2}{d-x} $$

Substitute \( x = \frac{2d}{3} \) and \( d-x = \frac{d}{3} \):

$$ V = \frac{k q_1}{(2d/3)} + \frac{k q_2}{(d/3)} = \frac{3k}{d} \left( \frac{q_1}{2} + q_2 \right) $$

4. Solving for Separation ‘d’

We substitute the known values into the equation. Use \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

$$ V = 900 \, \text{V} $$ $$ q_1 = 4 \times 10^{-9} \, \text{C}, \quad q_2 = 1 \times 10^{-9} \, \text{C} $$ $$ 900 = \frac{3 \times (9 \times 10^9)}{d} \left( \frac{4 \times 10^{-9}}{2} + 1 \times 10^{-9} \right) $$ $$ 900 = \frac{27 \times 10^9}{d} \times (2 \times 10^{-9} + 1 \times 10^{-9}) $$ $$ 900 = \frac{27 \times 10^9}{d} \times (3 \times 10^{-9}) $$ $$ 900 = \frac{81}{d} $$ $$ d = \frac{81}{900} = \frac{9}{100} \, \text{m} $$